Question Number 36544 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$$\int\frac{{dx}}{{tanx}+{cotx}+{secx}+{cosecx}} \\ $$
Answered by ajfour last updated on 03/Jun/18
$${I}=\int\frac{\mathrm{sin}\:{x}\mathrm{cos}\:{xdx}}{\mathrm{1}+\mathrm{sin}\:{x}+\mathrm{cos}\:{x}} \\ $$$$\:\:\:\:=\int\frac{\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\left(\mathrm{cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{sin}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)}{\mathrm{2cos}\:\frac{{x}}{\mathrm{2}}\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right)}{dx} \\ $$$$\:\:=\int\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\left(\mathrm{cos}\:\frac{{x}}{\mathrm{2}}−\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:{xdx}−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}−\mathrm{cos}\:{x}\right){dx} \\ $$$$\:\:\:=−\frac{\mathrm{cos}\:{x}}{\mathrm{2}}\:−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}+{c}\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$${excellent}... \\ $$
Commented by MJS last updated on 03/Jun/18
$$\mathrm{great}! \\ $$